Question

2

Answer 1

Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

moles of HCl= $molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol$

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water = $volume \times density=150.0ml\times 1.0g/ml=150.0g$

$q=m\times c\times \Delta T$

q = heat released

m = mass  = 150.0 g

c = specific heat = $4.184J/g^0C$

$\Delta T$ = change in temperature = $4.83^0C$

$q=150.0\times 4.184\times 4.83$

$q=3031.3J$

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = $\frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ$

Thus molar enthalpy change is 73.04 kJ/mol